Chemistry Q128 – AP-EAMCET 2025

Home Chemistry Redox ReactionsQuestion #551

In acid medium dichromate oxidizes sulphite to sulphate as shown below:

x {Cr}_{2} O_{7}^{2 -} + y {SO}_{3}^{2 -} + z H^{+} ⟶ a {Cr}^{3 +} + b {SO}_{4}^{2 -} + c H_{2} O

Identify correct statements about this balanced equation (only) I) Sum of

x

and

y

is 4 II) Sum of

a

and

c

is equals to

(3 + b)

III) Sum of

x, y

and

z

is 11

Clue 1 of 3: Redox Reaction Balancing

Identify the oxidation and reduction half-reactions in the given chemical equation.

🧠 Full Solution Path

Step 1: Identify Oxidation and Reduction Half-Reactions✦ Active

The given reaction involves dichromate ( ${Cr}_{2} O_{7}^{2 -}$ ) being reduced to chromium(III) ( ${Cr}^{3 +}$ ), and sulphite ( ${SO}_{3}^{2 -}$ ) being oxidized to sulphate ( ${SO}_{4}^{2 -}$ ). We need to balance these two half-reactions in an acidic medium.

Step 2: Balance the Oxidation Half-Reaction○ Expand

For the oxidation of sulphite to sulphate:

{SO}_{3}^{2 -} ⟶ {SO}_{4}^{2 -}

Balance oxygen atoms by adding $H_{2} O$ :

{SO}_{3}^{2 -} + H_{2} O ⟶ {SO}_{4}^{2 -}

Balance hydrogen atoms by adding $H^{+}$ :

{SO}_{3}^{2 -} + H_{2} O ⟶ {SO}_{4}^{2 -} + 2 H^{+}

Balance charge by adding electrons. The left side has a charge of $- 2$ , and the right side has a charge of $0$ . Add 2 electrons to the right side:

{SO}_{3}^{2 -} + H_{2} O ⟶ {SO}_{4}^{2 -} + 2 H^{+} + 2 e^{-} (Equation 1)

💡 Teacher's Secret Hint

Remember to add $H_{2} O$ to the side deficient in oxygen and $H^{+}$ to the side deficient in hydrogen when balancing in acidic medium.

Step 3: Balance the Reduction Half-Reaction○ Expand

For the reduction of dichromate to chromium(III):

{Cr}_{2} O_{7}^{2 -} ⟶ {Cr}^{3 +}

Balance chromium atoms:

{Cr}_{2} O_{7}^{2 -} ⟶ 2 {Cr}^{3 +}

Balance oxygen atoms by adding $H_{2} O$ :

{Cr}_{2} O_{7}^{2 -} ⟶ 2 {Cr}^{3 +} + 7 H_{2} O

Balance hydrogen atoms by adding $H^{+}$ :

{Cr}_{2} O_{7}^{2 -} + 14 H^{+} ⟶ 2 {Cr}^{3 +} + 7 H_{2} O

Balance charge by adding electrons. The left side has a charge of $(- 2 + 14) = + 12$ , and the right side has a charge of $2 (+ 3) = + 6$ . Add 6 electrons to the left side:

{Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} ⟶ 2 {Cr}^{3 +} + 7 H_{2} O (Equation 2)

Step 4: Combine Half-Reactions and Determine Coefficients○ Expand

To combine the half-reactions, the number of electrons must be equal. Multiply Equation 1 by 3 to get 6 electrons:

3 ({SO}_{3}^{2 -} + H_{2} O ⟶ {SO}_{4}^{2 -} + 2 H^{+} + 2 e^{-})

3 {SO}_{3}^{2 -} + 3 H_{2} O ⟶ 3 {SO}_{4}^{2 -} + 6 H^{+} + 6 e^{-} (Equation 1')

Now, add Equation 1' and Equation 2:

({Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-}) + (3 {SO}_{3}^{2 -} + 3 H_{2} O) ⟶ (2 {Cr}^{3 +} + 7 H_{2} O) + (3 {SO}_{4}^{2 -} + 6 H^{+} + 6 e^{-})

Cancel electrons and simplify $H^{+}$ and $H_{2} O$ on both sides:

{Cr}_{2} O_{7}^{2 -} + (14 - 6) H^{+} + 3 {SO}_{3}^{2 -} ⟶ 2 {Cr}^{3 +} + (7 - 3) H_{2} O + 3 {SO}_{4}^{2 -}

{Cr}_{2} O_{7}^{2 -} + 3 {SO}_{3}^{2 -} + 8 H^{+} ⟶ 2 {Cr}^{3 +} + 3 {SO}_{4}^{2 -} + 4 H_{2} O

Comparing this with the given equation $x {Cr}_{2} O_{7}^{2 -} + y {SO}_{3}^{2 -} + z H^{+} ⟶ a {Cr}^{3 +} + b {SO}_{4}^{2 -} + c H_{2} O$ , we find the coefficients: $x = 1$ $y = 3$ $z = 8$ $a = 2$ $b = 3$ $c = 4$

💡 Teacher's Secret Hint

Always double-check the final balanced equation for atom and charge balance.

Step 5: Evaluate the Statements○ Expand

Now, let's check each statement: I) Sum of $x$ and $y$ is 4: $x + y = 1 + 3 = 4$ . This statement is **True**. II) Sum of $a$ and $c$ is equals to $(3 + b)$ : $a + c = 2 + 4 = 6$ . $3 + b = 3 + 3 = 6$ . Since $6 = 6$ , this statement is **True**. III) Sum of $x, y$ and $z$ is 11: $x + y + z = 1 + 3 + 8 = 12$ . Since $12 \neq 11$ , this statement is **False**.

Therefore, only statements I and II are correct.

💡 Teacher's Secret Hint

Carefully substitute the derived coefficients into each statement to avoid calculation errors.

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Chemistry Question 128 – AP-EAMCET 2025

Clue 1 of 3: Redox Reaction Balancing

🧠 Full Solution Path